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Anastasy [175]
3 years ago
14

1. The choice of materials for an exciting playground slide should __________.

Physics
2 answers:
gavmur [86]3 years ago
4 0
  1. have a large coefficient of kinetic friction between cloth and the slide material
  2. 40(0.3+0.5)=40(0.8)=32N
  3. F-applied=Wbook/μk
  4. When the tire is locked, the frictional force is greater but the car cannot be steered
Ugo [173]3 years ago
3 0

Answer:

1 have a small coefficient of kinetic friction between cloth and the slide material

2 12 N

3 F-applied=Wbook/μs

4 The static friction force is larger than the kinetic friction force

Explanation:

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Can A positively charged body attract another positively charged body​
andriy [413]

Like charges repel, unlike charges attract

Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

So I would say no, unless the two bodies are placed close to each other where one has much more charge than the other, then due to induction, force of attraction becomes more than the force of repulsion.

3 0
3 years ago
An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,
cupoosta [38]

First overtone of open organ pipe is given as

f_{1o} = \frac{v}{L_1 + 2e}

first overtone of closed organ pipe is given as

f_{1c} = \frac{3v}{4(L_2 + e)}

now they are in unison so we will have

\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}

\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}

90 + 6e = 92 + 4e

e = 1 cm

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8 0
3 years ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

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viktelen [127]
It would take roughly 3 seconds.
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